so i've got (|x| / 1+|x|) < 1
Is that enough of a description? Not sure what I could say about Br(a)

Maybe you can try to simplify !
1+|x|>0 for any x. So you can multiply both sides by 1+|x|, without changing the inequality :
which gives , for any x. It means that the inequality is true for any x.
So the open ball is the whole set.

And (X,d) a metric space, fix a point o in X
let d1(x,y)= d(x,o) + d(o,y)
if x ≠ y and d1(x,x)=0